(a) Use Newton’s method with x1 = 1 to find the root of the equation x3 − x = 1 correct to six decimal places.
(b) Solve the equation in part (a) using x1 = 0.6 as the initial approximation.
(c) Solve the equation in part (a) using x1 = 0.57.
(You definitely need a programmable calculator for this part.)
(d) Graph f(x) = x3 − x − 1 and its tangent lines at x1 = 1, 0.6, and 0.57 to explain why Newton’s method is so sensitive to the value of the initial approximation.
(a)
To find: The root of the equation
Answer to Problem 26E
The root is 1.324718.
Explanation of Solution
Formula used:
Newton’s method
Given:
Calculation:
Differentiate with respect to
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Hence the root is 1.324718
(b)
To find: The root of the equation
Answer to Problem 26E
The root is 1.324718.
Explanation of Solution
Given:
Formula used:
Newton’s method
Calculation:
Differentiate with respect to
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Hence the root is 1.324718
(c)
To find: The root of the equation
Answer to Problem 26E
The root is 1.324718
Explanation of Solution
Given:
Formula used:
Newton’s method
Calculation:
Differentiate with respect to
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Here,
Thus, the root of x is 1.324718.
(d)
To sketch: the graph
Explanation of Solution
Given:
Calculation:
The graph is,
In Figure 1,
The tangent line corresponding to
The tangent line corresponding to
The tangent line corresponding to
Chapter 4 Solutions
Single Variable Calculus: Concepts and Contexts, Enhanced Edition
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- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning